Elliptical Orbits and Angular Momentum

AI Thread Summary
The discussion centers on the expression for angular momentum in elliptical orbits, specifically questioning the absence of a sine term in the formula l = mr²φ̇. Participants clarify that while the initial assumption includes a sine term due to the angle between the radius and velocity vectors, the correct formulation simplifies this to l = mr²φ̇ without the sine. The conversation emphasizes the importance of using polar coordinates and understanding the relationship between radial and transverse components of motion. Participants recommend consulting standard classical mechanics texts for a deeper understanding of these concepts. The discussion highlights the nuances of angular momentum calculations in non-circular orbits.
WWCY
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Homework Statement



Why is the magnitude of Angular Momentum for an elliptical orbit as such?
$$l = mr^2\dot{\phi}$$
where ##\dot{\phi}## represents angular momentum.

I have always assumed that angular momentum was $$l = r \times P = mr \times V = mrVsin(\theta) = mr^2\dot{\phi} sin(\theta)$$
And since the angle (taken to be ##\theta##) between the ##r## and ##V## vectors isn't always a right angle, shouldn't angular momentum for elliptical orbits be defined with the sine term?

Assistance is greatly appreciated!

Homework Equations

The Attempt at a Solution

 
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WWCY said:

Homework Statement



Why is the magnitude of Angular Momentum for an elliptical orbit as such?
$$l = mr^2\dot{\phi}$$
where ##\dot{\phi}## represents angular momentum.

I have always assumed that angular momentum was $$l = r \times P = mr \times V = mrVsin(\theta) = mr^2\dot{\phi} sin(\theta)$$
And since the angle (taken to be ##\theta##) between the ##r## and ##V## vectors isn't always a right angle, shouldn't angular momentum for elliptical orbits be defined with the sine term?

Assistance is greatly appreciated!

Homework Equations

The Attempt at a Solution

The correct expression is
$${\mathbf V} = \dot{r}\, {\mathbf e}_r + r \dot{\phi} \, {\mathbf e}_{\phi}, $$
so
$$ {\mathbf r \times V} = r \dot{\phi}\, {\mathbf e}_r \times {\mathbf e}_{\phi}$$

Here, ##r = |{\mathbf r}|,## and ##{\mathbf e}_r##, ##{\mathbf e}_{\phi}## are the unit vectors in the ##r## and ##\phi## directions (so ##{\mathbf e}_r \times {\mathbf e}_{\phi} = {\mathbf e}_z##).

Be very careful to write vectors when you should have vectors, and scalars when you should have scalars.
 
Last edited:
WWCY, your expression is correct but not very useful. Depending on the path of the particle, θ could be a very complicated function of r and ∅. Ray Vickson's expression is always correct.
 
Ray Vickson said:
The correct expression is
$${\mathbf V} = \dot{r}\, {\mathbf e}_r + r \dot{\phi} \, {\mathbf e}_{\phi}, $$

Could you elaborate on the meaning of this expression? I don't think I've seen something similar before.

Ray Vickson said:
$$ {\mathbf r \times V} = r \dot{\phi}\, {\mathbf e}_r \times {\mathbf e}_{\phi}$$

How do we arrive at this expression? Also, is there not a ##r^2## in the final term for magnitude?

Thanks both for the replies. Apologies if this is an elementary question.
 
WWCY said:
l=mr2˙ϕ
There is (should be) an r2 in Ray Vickson's expression. The theory involves the use of polar coordinates and corresponding unit vectors (radial and transverse coordinates). If you are not familiar with that, then how did you arrive at your original expression for the angular momentum, which I quoted above?
 
Chandra Prayaga said:
There is (should be) an r2 in Ray Vickson's expression. The theory involves the use of polar coordinates and corresponding unit vectors (radial and transverse coordinates). If you are not familiar with that, then how did you arrive at your original expression for the angular momentum, which I quoted above?

Yes, of course ##{\mathbf l} = r^2 \dot{\phi} \, {\mathbf e}_z##.
 
Chandra Prayaga said:
There is (should be) an r2 in Ray Vickson's expression. The theory involves the use of polar coordinates and corresponding unit vectors (radial and transverse coordinates). If you are not familiar with that, then how did you arrive at your original expression for the angular momentum, which I quoted above?
My first exposure to angular momentum came from introductory texts that did not use such a theory.

Do you mind guiding me through the steps?

Thank you!
 
That treatment is available in any standard textbook on Classical Mechanics, such as the one by Taylor. It takes some time to work through it.
 
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